COMEDK 2024 Shift 2 Paper

To address the problem, we need to verify each option based on the given conditions and probability rules:
Given:
1. P(A∕B)=‌

1

2

2. P(B∕A)=‌

1

3

3. P(A∩B)=‌

1

6

We start by recalling that conditional probability is defined as:

‌P(A∕B)=‌

P(A∩B)

P(B)

‌P(B∕A)=‌

P(B∩A)

P(A)

=‌

P(A∩B)

P(A)

Now we'll use the given values to find P(A) and P(B).
From P(A∕B)=‌

1

2

‌‌

P(A∩B)

P(B)

=‌

1

2

‌‌

‌ 1 6

P(B)

=‌

1

2

‌P(B)=‌

‌ 1 6

1 2

=‌

1

3

From P(B∕A)=‌

1

3

:
‌‌

P(A∩B)

P(A)

=‌

1

3

‌‌

‌ 1 6

P(A)

=‌

1

3

‌P(A)=‌

‌ 1 6

1 3

=‌

1

2

Next, we will use these probabilities to check each option.
Option A:A and B are not independent
Events A and B are independent if:
‌P(A∩B)=P(A)P(B)
‌P(A)P(B)=‌

1

2

⋅‌

1

3

=‌

1

6

Since this equals P(A∩B), A and B are actually independent, making Option A false. Therefore, Option A is not true.
Option B: P(A∪B)=‌

2

3

Using the formula for the union of two events:
Therefore, Option B is true.
Option C: P(A′∩B)=‌

1

6

This requires calculating P(A′∩B) :
‌P(A′∩B)=P(B)−P(A∩B)
‌P(A′∩B)=‌

1

3

−‌

1

6

=‌

2

6

−‌

1

6

=‌

1

6

So, Option C is true.
Option D: A and B are independent
As we established earlier, A and B are independent because P(A∩B)=P(A)P(B). Thus, Option D is also true.
In summary, the option that is not true is:
Option A:A and B are not independent.